Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. \(9x^2+30x+25=\left(3x+5\right)^2\)
b. \(\dfrac{4}{9}x^4-16x^2=\left(\dfrac{2}{3}x^2-4x\right)\left(\dfrac{2}{3}x^2+4x\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)
c. \(a^2y^2+b^2x^2-2axby=\left(ay-bx\right)^2\)
d. \(100-\left(3x-y\right)^2=\left(10-3x+y\right)\left(10+3x-y\right)\)
e. \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)
f. \(64x^2-\left(8a+b\right)^2=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
g. \(27x^3-a^3b^3=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)
a) 9x2+30x+25=32x2+2.3.5x+52=(3x+5)2
b)12/5x2y2-9x4-4/25y4=-(9x4-12/5x2y2+4/25y4)=-(3x-2/5y)2
c)a2y2+b2x22axby=(ax-by)2
d)64x2-(8a+b)2=(8x-8a-b)(8x+8a+b)
a) = (3x)^2 + 2.3x.5+ 5^2 = (3x+5)^2
b) = (2/3x)^2-(4y)^2=(2/3x-4y)(2/3x+4y)
c) = -(9x^4-12/5x^2y^2+4/25y^4) = -[(3x^2)^2 - 2.3x^2.2/5y^2 + (2/5y^2)^2]= -(3x^2-2/5y^2)^2
d) = (x-5)^2 - 4^2= (x-5+4)(x-5-4) = (x-1)(x-9)
e) = (2x)^3 + 3.(2x)^2.(5y) + 3.(2x).(5y)^2 + (5y)^3 = (2x+5y)^3
f) = (8x)^2 - (8a+b)^2 = (8x-8a-b)(8x+8a+b)
g) = (7x-4-2x-1)(7x-4+2x+1) = (5x-5)(9x-3) = 5(x-1).3(x-3)=15(x-1)(x-3)
h) = (x-y)(x+y)- 2(x+y) = (x+y)(x-y-2)
# Chúc bạn học tốt #
b: \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
c: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
e: \(a^2y^2-2axby+b^2x^2\)
\(=\left(ay\right)^2-2\cdot ay\cdot bx+\left(bx\right)^2\)
\(=\left(ay-bx\right)^2\)
f: \(100-\left(3x-y\right)^2\)
\(=\left(10-3x+y\right)\left(10+3x-y\right)\)
g: \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
a, bằng cách tìm nhân tử chung
1,\(x^2-3x\)
=x.(\(\left(x-3\right)\)
2,\(15x^2-6x\)
=3x.(5x-2)
3,\(4x\left(x-y\right)\)\(+2y\left(x-y\right)\)
=(x-y).(4x+2y)
=2(x-y).(x+y)
=2(\(x^2-y^2\left(\right)\)
b, dùng hằng đẳng thức
1,\(64x^2-25y^2\)
=\(\left(8x\right)^2-\left(5y\right)^2\)
=(8x-5y)(8x+5y)
2,\(9x^2-30x-25\)
=\(\left(3x-5\right)^2\)
3,
\(\dfrac{1}{4}x^2+2x+4\)
=\(\left(\dfrac{1}{2}x+2\right)^2\)
4,\(25a^2-2a+\dfrac{1}{25}\)
=(\(\left(5a-\dfrac{1}{5}\right)^2\)
Làm bài 1 thôi !! Mấy bài kia tương tự . Tìm nhân tử chung ra .
a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2=\left(x^2+x-1+x^2+2x+3\right)\left(x^2+x-1-x^2-2x-3\right)\)
\(=\left(2x^2+3x+2\right)\left(-x-4\right)\)
c) \(-16+\left(x-3\right)^2=\left(x-3+4\right)\left(x-3-4\right)=x\left(x-7\right)\)
d) \(64+16y+y^2=\left(y+8\right)\left(y+8\right)\)
a , \(16x^2+8x+1=\left(4x\right)^2+2.4x.1+1^2=\left(4x+1\right)^2\)
b , \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
a,(4x+1)2 e,\(\left(\dfrac{3}{2}x-\dfrac{2}{5}\right)^2\)
b,(x-\(\dfrac{1}{2}\))2 g,\(\left(xy+1\right)^2\)
c,(\(x+\dfrac{3}{2}\))2 h,\(\left(x+5\right)^2\)
d,\(\left(x-\dfrac{5}{4}\right)^2\) i,\(-\left(x-6\right)^2\)
k,\(-\left(2x+3\right)^2\)
a: \(9x^2+30x+25=\left(3x+5\right)^2\)
b: \(\dfrac{4}{9}x^4-16x^2=x^2\left(\dfrac{4}{9}x^2-16\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)
c: \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4\)
\(=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)\)
\(=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)